GATE 2022 EC Q.39: solution and answer

GATE 2022 · ECQ.39 · 2 marks · MCQ

Consider the circuit shown in the figure with input $V (t)$ in volts. The sinusoidal steady state current $I (t)$ flowing through the circuit is shown graphically (where $t$ is in seconds). The circuit element $Z$ can be ________.

GATE 2022 EC question 39: Left: a single series loop. — Figure from the GATE 2022 EC paper. It is part of the question.

(A)
a capacitor of 1 F
(B)
an inductor of 1 H
(C)
a capacitor of $3$ F
(D)
an inductor of $3$ H

Answer and solution

ANSWER

The answer is (B): Z is an inductor of 1 H. The current's peak is 1/√2 A and it lags the 1 V-peak source by 45°. So Z_total = √2∠45° = 1 + j1 Ω, the 1 Ω resistor supplies the real part, and Z = j1 Ω = jωL with ω = 1 rad/s.

Official answer: BSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Source phasor. $V (t) = sin t$ gives $ω = 1 rad/s$ and a peak of $V_{m} = 1 V$ . Taking $sin$ as the phase reference, the rms phasor is $V = \frac{1}{2} ∠ 0^{\circ} V$ (rms).
Amplitude from the plot. The current just touches the dotted lines at $\pm \frac{1}{2} A$ , so its peak is $I_{m} = \frac{1}{2} A$ and its rms value is $\frac{I _{m}}{2} = \frac{1}{2} A$ . Then $∣ Z_{t o t a l} ∣ = \frac{V _{m}}{I _{m}} = \frac{1}{1/ 2} = 2 Ω$ . The ratio of the rms values gives the same answer.
Phase from the plot. Write $I (t) = I_{m} sin (t + ϕ)$ . This crosses zero going upward when $t + ϕ$ is a multiple of $2 π$ . The plot puts a rising zero crossing at $t = π /4$ , so $ϕ = - π /4$ (up to a multiple of $2 π$ ). That gives $I (t) = \frac{1}{2} sin (t - \frac{π}{4}) A$ . The current lags the voltage by $4 5^{\circ}$ : $I = \frac{1}{2} ∠ - 4 5^{\circ} A$ (rms).
Total impedance from Ohm's law in phasor form. $Z_{t o t a l} = \frac{V}{I} = \frac{\frac{1}{2} ∠ 0 ^{\circ}}{\frac{1}{2} ∠ - 4 5 ^{\circ}} = 2 ∠4 5^{\circ} = 2 (cos 4 5^{\circ} + j sin 4 5^{\circ}) = 1 + j 1 Ω$
Series combination. The $1 Ω$ resistor and $Z$ are in series, so $Z_{t o t a l} = 1 + Z$ . That gives $Z = j 1 Ω$ . This is a purely reactive impedance with positive reactance, so $Z$ is an inductor.
Element value. $j ω L = j 1$ with $ω = 1 rad/s$ gives $L = 1 H$ . Check: $∣1 + j 1∣ = 2 Ω$ and the lag is $tan^{- 1} (ω L / R) = tan^{- 1} (1) = 4 5^{\circ}$ . Both match the plot. The other options fail. A $1 F$ capacitor gives $Z_{t o t a l} = 1 - j 1 Ω$ , so the current would lead by $4 5^{\circ}$ and its rising zero crossing would be at $t = - π /4$ . Either $3$ element gives $∣ Z_{t o t a l} ∣ = 1 + 3 = 2 Ω$ , so the peak current would be $\frac{1}{2} A$ , not $\frac{1}{2} A$ . Answer: option (B), an inductor of 1 H.

KEY IDEA

In a series circuit driven by a sinusoid, the source phasor divided by the current phasor is the total impedance. The two readings come straight off the plot. The ratio of peak voltage to peak current gives |Z_total|. The labelled time of the current's rising zero crossing, multiplied by ω, gives the angle. A rising zero crossing at a positive time means the current lags, so the circuit is inductive. Subtract the known resistor, and what is left is jωL (inductor) or −j/(ωC) (capacitor).

WATCH OUT FOR

Reading the rising zero crossing at t = +π/4 as a lead. The current reaches that point of its cycle π/4 s later than the source does, so it lags. Calling it a lead gives the 1 F capacitor (A). A second trap is mixing peak and rms values. 1/√2 A on the plot is the current's PEAK. Dividing the source's rms value (1/√2 V) by that peak gives |Z_total| = 1 Ω, which leaves nothing for Z and matches no option. A third slip is forgetting that the reactance is ωL. Here ω = 1 rad/s, so X = 1 Ω means L = 1 H, but that only works because ω happens to be 1.

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 34 AC, power & resonance questions, with notes

GATE 2022 EC Q.39: solution and answer

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